From the following data at 25-C- The standard enthalpy of formation of FeOs is X- then 0-3-X is-Reaction - rH- kJ - mole Fe2O3s-3C graphite- 2Fes-3COg492-6 FeOs-C graphite- Fes-COg155-8 C graphite-O2g- CO2g -393-51 COg-1-2 O2g- CO2g -282-98

FeO( s ) +C( graphite ) ⟶ Fe( s ) +CO( g ) , 155.8 CO( g ) + 1 2 O _ 2 ( g ) ⟶ C O _ 2 ( g ) , 282.98 + FeO( s ) +C( graphite ) + 1 2 O _ ...

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Hess' Law

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Question

From the following data at $25^{\circ} C$ , The standard enthalpy of formation of FeO(s) is X, then 0.3-X is:

Reaction $Δ_{r} H^{\circ}$ ( kJ / mole )
$F e_{2} O_{3} (s) + 3 C (g r a p h i t e) \to 2 F e (s) + 3 C O (g)$ 492.6
$F e O (s) + C (g r a p h i t e) \to F e (s) + C O (g)$ 155.8
$C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g)$ $-$ 393.51
$C O (g) + 1 / 2 O_{2} (g) \to C O_{2} (g)$ $-$ 282.98

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{F e}_{2} O_{3} (s)

25^{o} C

{F e}_{2} O_{3} (s) + 3 C_{g r a p h i t e} ⟶ 2 F e (s) + 3 C O (g); Δ H^{o} = 492 k J / m o l

F e O (s) + C_{g r a p h i t e} ⟶ F e (s) + C O (g); Δ H^{o} = 155.8 k J / m o l

C_{g r a p h i t e} + O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 393.51 k J / m o l

C O (g) + \frac{1}{2} O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 282.98 k J / m o l

F e O (s)

{F e}_{2} O_{3} (s)

F e_{2} O_{3 (s)} + 3 C O_{(g)} \to 2 F e (s) + 3 C O_{2 (g)}

Δ H = - 26.8 k J

F e O_{(s)} + C O_{(g)} \to F e_{(s)} + C O_{2 (g)} Δ H = - 16.5 k J

Δ H

F e_{2} O_{3 (s)} + C O_{(g)} \to 2 F e O_{(s)} + C O_{2 (g)}

C H_{3} O H (l)

C H_{3} O H (l) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)

Δ H^{o} = - 726

m o l^{- 1}

+ O_{2} (g) \to C O_{2} (g);

Δ_{c} H^{0} = - 393

m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l);

Δ_{t} H^{o} = - 286

m o l^{- 1}

F e O (s) + C (g r a p h i t e) ⟶ F e (s) + C O (g)

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